Answer:
r=42227Km using 24h, r=42150Km using the exact given value.
Explanation:
The force that acts on the satellite of mass m is the gravitational pull of the Earth, of mass M. If the distance between their centers is r, we know that this gravitational force must be:
[tex]F_G=\frac{GMm}{r^2}[/tex]
Where [tex]G=6.67\times10^{-11}m^3/Kgs^2[/tex] is the gravitational constant.
The satellite moves in a circular trajectory because the net forces acting on it are centripetal, so we write the equation of the centripetal force:
[tex]F_{cp}=\frac{mv^2}{r}[/tex]
Since only the gravitational force is acting on the satellite this force is the net force, and thus, equal to the centripetal force:
[tex]F_G=F_{cp}[/tex]
Which means:
[tex]\frac{GMm}{r^2}=\frac{mv^2}{r}[/tex]
Or:
[tex]\frac{GM}{r}=v^2[/tex]
The velocity of the satellite is [tex]v=C/t[/tex], where C is the circumference of the orbit, whose radius is obviously r: [tex]C=2\pi r[/tex], so we can write:
[tex]\frac{GM}{r}=(\frac{2\pi r}{t})^2=\frac{4\pi^2 r^2}{t^2}[/tex]
Which means:
[tex]r^3=\frac{GMt^2}{4\pi^2}[/tex]
Which is Kepler's 3rd Law for a circular motion. We can write this as:
[tex]r=\sqrt[3]{\frac{GMt^2}{4\pi^2}}[/tex]
Since there are 60 seconds in a minute and 60 minutes in an hour, using 24 hours we have:
[tex]r=\sqrt[3]{\frac{(6.67\times10^{-11}m^3/Kgs^2)(5.97\times10^{24})(24\times60\times60s)^2}{4\pi^2}}=42226910m=42227Km[/tex]
We could use the exact time of (23)(60)(60)+(56)(60)+(4.1) seconds, and in that case we would obtain r=42150Km
